1. **Problem 10:** Given bearings: B from A is 070°, C from A is 120°, and C from B is 125°. (a) Find the bearing of A from C. (b) Find the bearing of B from C. 2. **Problem 11:** Two runners start from A. - Deepika runs 8 km on bearing 095° to B. - Khadeeja runs 8 km on bearing 155° to C. (a) Find distance BC. (b) Find bearing of B from C. (c) Find bearing of C from B. 3. **Problem 12:** B is 5 km from A on bearing 038°. C is 12 km from B on bearing 128°. Find distance AC. --- ### Step-by-step solutions: **10(a) Bearing of A from C:** 1. Bearings are clockwise from North. 2. Bearing of C from A is 120°, so line AC points 120° from North. 3. Bearing of A from C is opposite direction: add 180° to 120°. 4. $$120° + 180° = 300°$$ 5. Bearing must be between 0° and 360°, so bearing of A from C is **300°**. **10(b) Bearing of B from C:** 1. Bearing of C from B is 125°. 2. Bearing of B from C is opposite direction: $$125° + 180° = 305°$$ 3. Since 305° < 360°, bearing of B from C is **305°**. --- **11(a) Distance BC:** 1. Use Law of Cosines. Angle between bearings 095° and 155° is $$155° - 95° = 60°$$. 2. Both runners run 8 km. 3. Law of Cosines: $$BC^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos(60°)$$ 4. $$BC^2 = 64 + 64 - 128 \times 0.5 = 128 - 64 = 64$$ 5. $$BC = \sqrt{64} = 8$$ km. **11(b) Bearing of B from C:** 1. Use Law of Sines to find angle at C. 2. Let angle at C be $$\theta$$. 3. $$\frac{\sin(\theta)}{8} = \frac{\sin(60°)}{8}$$ so $$\sin(\theta) = \sin(60°) = 0.866$$ 4. $$\theta = 60°$$ (since triangle is isosceles) 5. Bearing of B from C = bearing of C from A + 180° - $$\theta$$ 6. $$155° + 180° - 60° = 275°$$ 7. Bearing of B from C is **275°**. **11(c) Bearing of C from B:** 1. Given as 125°. --- **12. Distance AC:** 1. B is 5 km from A at 038°. 2. C is 12 km from B at 128°. 3. Find angle ABC: - Bearing B from A is 038° - Bearing C from B is 128° - Angle between lines AB and BC is $$128° - 38° = 90°$$ 4. Use Law of Cosines: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(90°)$$ 5. $$AC^2 = 5^2 + 12^2 - 2 \times 5 \times 12 \times 0 = 25 + 144 = 169$$ 6. $$AC = \sqrt{169} = 13$$ km. --- **Final answers:** - 10(a) Bearing of A from C = **300°** - 10(b) Bearing of B from C = **305°** - 11(a) Distance BC = **8 km** - 11(b) Bearing of B from C = **275°** - 11(c) Bearing of C from B = **125°** - 12 Distance AC = **13 km**